Problem: $f(x, y) = \left( \cos(xy), 1 \right)$ $\text{div}(f) = $
Solution: The formula for divergence in two dimensions is $\text{div}(f) = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial P}{\partial x} &= \dfrac{\partial}{\partial x} \left[ \cos(xy) \right] \\ \\ &= -y\sin(xy) \\ \\ \dfrac{\partial Q}{\partial y} &= \dfrac{\partial}{\partial y} \left[ 1 \right] \\ \\ &= 0 \end{aligned}$ Adding the two partial derivatives, $\text{div}(f) = -y\sin(xy)$.